Question

Решите неравенства: 4<3x+1 <=10

Answer 1

$3 < 3x \leqslant 9 \\ 1 < x \leqslant 3$

Answer 2

To solve the compound inequality 4 < 3x + 1 ≤ 10, we need to break it down into two separate inequalities and solve them individually.

First, let's solve for the left inequality: 4 < 3x + 1.

Step 1: Subtract 1 from both sides of the inequality: 4 - 1 < 3x + 1 - 1 3 < 3x

Step 2: Divide both sides by 3 (note that since the coefficient of x is positive, the inequality sign remains the same): 3/3 < 3x/3 1 < x

Now, let's solve for the right inequality: 3x + 1 ≤ 10.

Step 1: Subtract 1 from both sides of the inequality: 3x + 1 - 1 ≤ 10 - 1 3x ≤ 9

Step 2: Divide both sides by 3: (3x)/3 ≤ 9/3 x ≤ 3

So, we have two inequalities:

1 < x (left inequality) x ≤ 3 (right inequality)

To find the solution for the original compound inequality, we need to find the intersection of the solutions for both inequalities. The overlapping interval is the final solution.

The solution is: 1 < x ≤ 3.